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### Topic: Acid/Base Questions  (Read 5184 times)

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#### luqman

• Guest ##### Acid/Base Questions
« on: May 12, 2004, 02:42:31 PM »
Hey guys,

Can someone plz check if this is correct, if not plz advise me on the correction.

Question 1;-

10.0cm3 portions of barium hydroxide solution, Ba(OH)2 (aq) of unknown concentration, were titrated with hydrochloric acid of concentration 0.100moldm-3.
16.2cm3 of hydrochloric acid was required for complete neutralization.

a) Write the balanced equation, with state symbols, for the reaction between barium hydroxide and hydrochloric acid to give barium chloride solution and water.

b) Calculate the concentration, in mol dm-3 of the barium hydroxide solution.

a) Ba (OH)2(aq) + 2HCl(aq) ? BaCl2 + 2H2O (l)

(b) Number of moles = 0.100 x 16.2             0.0162mol of HCL
1000
Ba(OH)2 : HCl
2  : 1

0.00162/2 = 0.00081 moles of Ba(OH)2

0.00081 x 1000        =     0.081moldm-3
/ 10.0

Question 2;-

A solution contained sodium hydroxide which was reacting slowly with other things present. In order to find the amount of sodium hydroxide present at a particular time, 100.0cm3 of 1.00moldm-3 hydrochloric acid (an excess) was added to neutralise the sodium hydroxide and stop the reaction. The mixture was made up to exactly 250.0cm3 with pure water. A 25.0cm3 sample of diluted mixture was titrated with 0.100moldm-3 sodium hydrogencarbonate solution. 40.0cm3 was required to prodcue a neutral solution.

a) Write a balanced equation, with state symbols, for the reaction of sodium hydroxide with hydrochloric acid.

b) Write a balanced equation for the reaction of sodium hydrogencarbonate with hydrochloric acid.

c) Calculate the number of moles of sodium hydroxide in the original solution.

(a) NaOH(aq) + HCl(aq)          NaCl(aq) + H2O(l)

(b) NaHCO3(aq) + HCL(aq) ? NaCl(aq) + H2O(l) + CO2(g)

Number of moles of NaHCO3 in titration

Number of moles = Molarity × Volume
1000

Number of moles  = 25.0 cm3 × 0.100 moldm-3    =  0.0025moldm-3
/ 1000

Ratio of NaHCO3 : HCl   1:1

Number of moles = Molarity × Volume        100.0 x 1.00   =   0.1 moldm-3
/ 1000                          /  1000
« Last Edit: May 14, 2004, 11:24:53 PM by hmx9123 »